Why is a critical point only a candidate for an extremum?
You will be able to: Identify domain-valid critical numbers and separate local from global comparisons.
BC foundation: Unit 5 shares these analytical differentiation objectives with AB. Check theorem hypotheses, justify derivative signs over intervals, and compare feasible candidates before claiming an optimum. State the local branch when analyzing an implicit relation.
Why is a critical point only a candidate for an extremum?
A hiking path can have a small hilltop lower than a distant peak. A local high point wins only against nearby heights; the absolute maximum wins against every permitted height.
A useful starting point: When are highest and lowest values guaranteed to exist? →
Words and symbols before equations
- Critical number c
- An interior domain input where f′(c)=0 or f′(c) does not exist.
- Critical point
- The graph point (c,f(c)) at a critical number.
- Local extremum
- A high or low compared with sufficiently nearby domain points.
- Stationary point
- A point with derivative zero.
What this picture assumes
Original model; numerical readouts are rounded. Local extrema here mean interior extrema. Zero is in the domain of the first three examples; it is excluded for 1/x. Reciprocal curves are split at zero.
Read the picture in three steps
- Read the axes and labels first. Identify what each symbol and line represents. Read the units and fixed conditions before comparing quantities.
- x²: zero is stationary and a minimum; f′ changes negative to positive.
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the mathematics
An interior local extremum must be a critical point. If the derivative exists there, a positive or negative slope would allow nearby values on opposite sides of the current height; a differentiable extremum therefore needs slope zero.
The converse fails: f=x³ has f′(0)=0, but keeps increasing through zero, so it has no local extremum there. A critical point is a candidate, not a verdict.
For f=abs(x), the derivative fails at 0 while the function is defined; this corner is a local and absolute minimum. For f=1/x, zero is not in the domain and is not a critical number.
In these lessons local extrema refer to interior points. Endpoints are tested separately for absolute extrema. Some texts use one-sided local-endpoint conventions; the candidate comparison itself is unchanged.
| Feature | Local extremum | Absolute extremum |
|---|---|---|
| Comparison | Sufficiently nearby inputs | Every input in the stated domain |
| Endpoints here | Handled separately | Included when in the domain |
| Candidate is enough? | No: classify it | No: compare all eligible values |
A worked example, step by step
Classify the critical point at zero for x², x³ and abs(x).
- For x², f′=2x is zero at 0 and the function has a minimum there.
- For x³, f′=3x² is zero at 0 but there is no sign change from increasing to decreasing or vice versa.
- For abs(x), the derivative does not exist at 0, which is in the domain.
- The corner is a minimum; critical points may be smooth extrema, non-extrema or nonsmooth extrema.
Do not discard a point just because its derivative fails, and do not include a derivative singularity outside the function’s domain.
Is zero critical for 1/x?
Compare with an explanation
No. The function itself is undefined there.
Predict. Change one thing. Explain.
Switch among x², x³, abs(x) and 1/x. Decide whether zero is in the function’s domain, whether it is critical and whether it is an extremum.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
x²: zero is stationary and a minimum; f′ changes negative to positive.
Original model; numerical readouts are rounded. Local extrema here mean interior extrema. Zero is in the domain of the first three examples; it is excluded for 1/x. Reciprocal curves are split at zero.
Explain what you noticed: Answer the investigation prompt above. State one observation and explain it using the relevant derivative signs, theorem conditions, domain or geometric constraint. Identify what the representation cannot tell you.
Apply the idea to a fresh problem Practice →Show what you understand.
Two original questions are a starting check, not proof of mastery. Explain your choice before revealing the answer.
Original written challenge
4 points · self-check · not an official AP questionFor f=x³−3x, find the critical numbers, then state the local-extremum values after checking derivative signs.
This response is not submitted or saved. Copy it before leaving.
Compare with the answer and four-point rubric
- 1 point: f′=3x²−3=3(x−1)(x+1).
- 1 point: The critical numbers are −1 and 1.
- 1 point: The derivative changes + to − at −1 and − to + at 1.
- 1 point: Local maximum f(−1)=2; local minimum f(1)=−2.
Accept equivalent correct methods and explanations. This is a Refresh Kid teaching rubric, not an official AP scoring guideline.
Retrieve it before you reveal it.
RECALL 1Why inspect the original domain?
A critical number must be an input where the function exists.
RECALL 2Must critical points be extrema?
No.
RECALL 3What does local compare?
Nearby values rather than the entire domain.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Why is a critical point only a candidate for an extremum?
- Critical candidates: f′=0 or f′ undefined, with f defined.
- Interior local extrema require candidates; candidates need further testing.
Remember: Do not discard a point just because its derivative fails, and do not include a derivative singularity outside the function’s domain.
Conditions: Original model; numerical readouts are rounded. Local extrema here mean interior extrema. Zero is in the domain of the first three examples; it is excluded for 1/x. Reciprocal curves are split at zero.
Refresh Kid · AP Calculus BC Unit 5 · Objectives FUN-1.C · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 5.2, FUN-1.C. CED effective Fall 2020, with Fall 2026 clarifications, checked September 17, 2026. Unit 5 has twelve official topics. Focused lesson titles and questions are original Refresh Kid teaching material.
Existence theorems require their stated hypotheses. Critical numbers must belong to the function’s domain; interior local extrema and included endpoints are handled explicitly. Extrema and inflection candidates need justification, not just a zero derivative. Sparse derivative samples do not establish interval-wide signs. Optimization includes the constraint, feasible domain and global comparison. Implicit-curve conclusions specify the branch and distinguish finite slopes from vertical tangents.
The Organic Chemistry Tutor Mean Value Theorem and Optimization Problems video titles, creator and relevant descriptions were checked; full videos were not reviewed. The free optimization video mentions additional paid material, which is not required here. Khan Academy’s unit destination was checked; its JavaScript lesson contents were not fully readable by the research tool. OpenStax Sections 4.3, 4.4, 4.5 and 4.7 were consulted for conceptual cross-checking. No creator scripts, questions, diagrams or artwork were copied. Refresh Kid is not affiliated with or endorsed by these providers.
GitHub’s 3D website collection and its camera-control example informed optional spatial inspection. The open-box model is original geometry using existing self-hosted Three.js with its MIT license. A 12 cm square sheet loses four equal corner squares. The remaining net folds into an open box; the geometry uses the same lengths as the labeled 2D net. Fold angle is a construction view, not an independent design variable; displayed volume refers to the fully upright box. Camera rotation changes no mathematical values. Complete 2D diagrams and explanations remain available without WebGL.
Independent teacher review and observation of students remain pending. Technical checks do not certify mathematical accuracy, accessibility or learning effectiveness. This is a review edition.
Optional official resource: Released AP Calculus BC questions and scoring guides. The archive spans multiple units; no entire exam question is assigned here.
Learn → Explore → Practice → Review is informed by the IES learning guide; this exact implementation has not been evaluated with learners.
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